Let $E/F$ be finite separable extension. Is there any proof of the fact that there are only finitely many intermediate fields without using primitive element theorem or fundamental theorem of Galois theory?
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7$\begingroup$ What's the motivation for not wanting to use either of those results? Considering the primitive element theorem is often stated as a finite extension has a primitive element if and only if there are a finite number of intermediate fields. $\endgroup$– JSchlatherCommented Feb 7, 2013 at 14:48
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3$\begingroup$ If one cannot use galois-theory, then why tag this galois-theory? $\endgroup$– awllowerCommented Feb 7, 2013 at 15:00
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2$\begingroup$ Because I thought it is related to galois theory or can be used in galois theory. $\endgroup$– MohanCommented Feb 7, 2013 at 15:02
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1$\begingroup$ Indeed. Thanks for the clarification. $\endgroup$– awllowerCommented Feb 7, 2013 at 15:11
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5$\begingroup$ @Mohan: Lang in his Algebra shows directly that a finite separable extension has a primitive element, and thus your result follows using the (stronger form of) The Primitive Element Theorem but no Galois Theory. (This argument is replicated in $\S 8$ of my field theory notes: math.uga.edu/~pete/FieldTheory.pdf.) I think this is not what you are looking for. Could you amplify on why you want to avoid the Primitive Element Theorem and Galois Theory? $\endgroup$– Pete L. ClarkCommented Mar 3, 2013 at 17:57
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1 Answer
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I think of the following argumentation:
1) Let $K$ be a field and $n\in \mathbb{N}$. Then $K^n$ has only finite many commutative unital subalgebras. The exact number is $B(n)$ - the $n$th Bell number.
2) Let $(K;L)$ be a separable extension. Then $L$ is separable as $K$-algebra. Thus, by tensoring a suitable splitting field $T$ for $L$ we get that $L\otimes T$ is isomorphic to $T^n$ for a suitable $n$. Hence - using 1) - $L\otimes T$ has only finite many commutative unital subalgebras.
3) The function $A\mapsto A\otimes L$ is injective defined on the commutative unital subalgebras of $L$.
4) Any commutative unital subalgebra of $L$ is a field, too.
Remark: Such algebras are called futile (with only finite many subalgebras) in the literature.
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$\begingroup$ I don't think that 3) is correct. Let $K=\mathbb{Q}$, $L=\mathbb{Q}(\sqrt{2}, \sqrt{3})$, then $\mathbb{Q}(\sqrt{2}) \otimes \mathbb{Q}(\sqrt{2}, \sqrt{3}) \cong \mathbb{Q}(\sqrt{3}) \otimes \mathbb{Q}(\sqrt{2}, \sqrt{3})$. $\endgroup$– AlexCommented Oct 27, 2019 at 7:30